Deduce the following equality: $$\prod_{i=0}^{n-1} (3^2+2i) = \frac{2^{n+5} \Gamma(n+\frac{11}{2})}{105 \sqrt{\pi}}$$
So I have the product by expansion: $$\prod_{i=0}^{n-1} (3^2+2i)=3^2(3^2+2\cdot 1)(3^2+2 \cdot 2)(3^2+ \cdots)(3^2+(n-1) \cdot 2)$$
I prefer to put this into binomial form, when doing so $$\prod_{i=0}^{n-1} (3^2+2i) = \binom{3^2+(n-1) \cdot 2}{3^2} \\ = \binom{\frac{3^2}{2}+(n-1)}{\frac{3^2}{2}} $$
Then in $\Gamma$ representation, I have the following $$x\Gamma(x)=\Gamma(x+1) \\ \prod_{i=0}^{n-1} (3^2+2i) = \frac{\Gamma(\frac{3^2}{2}+n)}{(3^2/2-1)(3^2/2-2)(3^2/2-3)(3^2/2-4)\Gamma(\frac{1}{2})} \\ = \frac{2^4 \Gamma(\frac{3^2}{2}+n)}{105\Gamma(\frac{1}{2})}$$
However, I am not deriving the same result.
Making the problem more general $$\prod_{i=0}^{n-1} (a+b\,i)=\prod_{i=0}^{n-1} a\,\,\prod_{i=0}^{n-1} \left(1+\frac{b }{a}\,i\right)=a^n \,\,\,\prod_{i=0}^{n-1} \left(1+\frac{b }{a}\,i\right)$$
Using Pochhammer symbols and converting them in terms if the gamma function $$\prod_{i=0}^{n-1} \left(1+c\,i\right)=c^{n-1} \left(\frac{c+1}{c}\right)_{n-1}=c^n\frac{ \Gamma \left(n+\frac{1}{c}\right)}{\Gamma \left(\frac{1}{c}\right)}$$ Using $c=\frac ba$ $$\prod_{i=0}^{n-1} (a+b\,i)=b^{n-1}\,\,\frac{ \Gamma \left(n+\frac{a}{b}\right)}{\Gamma \left(1+\frac{a}{b}\right)}=b^n\,\,\frac{ \Gamma \left(n+\frac{a}{b}\right)}{\Gamma \left(\frac{a}{b}\right)}$$
For $a=9$ and $b=2$, you then have $$\prod_{i=0}^{n-1} (9+2\,i)=\frac{2^{n+4}\,\, \Gamma \left(n+\frac{9}{2}\right)}{105 \sqrt{\pi }}$$ which is not your result.
To obtain the rhs, you need $\large\prod_{i=0}^n$ and not $\large\prod_{i=0}^{n-1}$